TITLE: A bounded sequence whose sequence of averages does not converge
QUESTION [5 upvotes]: Can we find a bounded sequence $\{a_n\}$ such that the sequence of its averages, say, sequence $\{b_n\}$, where 
$$b_n=\frac{1}{n}\sum_{i=1}^n a_i,$$
does not converge?

REPLY [2 votes]: One way is to have a sequence of zeroes and ones. You add $1$'s to your sequence until the average goes above say ${3 \over 4}$, then add $0$'s to your sequence until the average goes below ${1 \over 4}$, then add more $1$'s until it goes above ${3 \over 4}$, and back to zeroes ad infinitum.
A useful fact in showing this works: If $n = n_1 + n_2$, then 
$${1 \over n} \sum_{k = 1}^n a_k = {n_1 \over n} ({1 \over n_1} \sum_{k = 1}^{n_1} a_k) + {n_2 \over n} ({1 \over n_2} \sum_{k = n_1 + 1}^{n} a_k)$$
So if $A_k$ denotes the average of the first $k$ members of your sequence and $B_{n_1,n}$ denotes the average of the $n_1 + 1$ through $n$th members  you have
$$A_n = {n_1 \over n} A_{n_1} + {n_2 \over n} B_{n_1,n}$$
$$= {n_1 \over n} A_{n_1} + (1 - {n_1 \over n}) B_{n_1,n}$$
So by increasing $n$ for a given $n_1$, your average $A_n$ becomes a weighted average of
$A_{n_1}$ and $B_{n_1,n}$ that emphasizes $B_{n_1,n}$ more and more.